GATE CE 2022 Official Paper: Shift 1 (Previous Year Paper)

The correct answer is 'know / no'.

Key Points 

• The given words are homophones i.e. similar sounding words.
• Know: be aware of through observation, inquiry, or information.
• For eg.- Most people know that CFCs can damage the ozone layer.
• No: not any.
• For eg.- There is no excuse.

The complete sentence will be: You should know when to say no.

• Hence, option 4 is the correct answer.

Additional Information

• Ozone: colourless unstable toxic gas with a pungent odour and powerful oxidizing properties, formed from oxygen by electrical discharges or ultraviolet light. It differs from normal oxygen (O2) in having three atoms in its molecule (O3).

Explanation-

Given that, two straight lines are passing through the origin (0,0).

One of the straight lines is passing through the point (x1, y1) = (1, 3) and the other passes through the point (x2, y2) = (1, 2).

The area enclosed in the interval (0,1) = Area of OAC - Area of OAB.

Area of OAC = 0.5 × 3 × 1 = 1.5 sq. unit

Area of OAB = 0.5 × 2 × 1 = 1 sq. unit

Area enclosed = 1.5 - 1 = 0.5 sq. unit

Explanation-

Given that,

p ∶ q = 1 ∶ 2 ------ (i)

q ∶ r = 4 ∶ 3  ------ (ii)

r ∶ s = 4 ∶ 5  ----- (iii)

There has to be a common number between Equations 1 and 2, so let's multiply 8 in equation 1 and 4 in equation 2.

Similarly, multiply 3 in equation 3.

So the new equations will be 

p ∶ q = 8 ∶ 16 ------ (iv)

q ∶ r = 16∶ 12  ------ (v)

r ∶ s = 12 ∶ 15  ----- (vi)

From equation iv, v, vi we have p : q : r : s = 8 : 16 : 12 : 15 

Let s = 15, u is 50% more than s so u = 1.5 × 15 = 22.5 

Ratio of p and u, p : u = 8 : 22.5 = 16 : 45

Explanation-

Family table-

Given that, P is the sister of Q

Q is the husband of R

R is the mother of S

T is the husband of P.

The family tree diagram will be, 

 

So from the above tree diagram, T is an "uncle" of S.

Explanation-

From the figure, for triangle PQR, using Pythagoras theorem, 

PR² = PQ² + QR²

From the circle we know, QR = RZ ( Radius of circle)

So, PR2 = PQ2 + RZ2

It is known that area of square = square of the side

So PR2 =  Area of square PXWR

PQ2 =  Area of square SPQT

RZ2 =  Area of square RUVZ

Area of square PXWR = Area of square SPQT + Area of square RUVZ

Area of SPQT = Area of PXWR - Area of RUVZ

The correct answer is 'Eating healthy can be started at any age, earlier the better'.

Key Points

• Let's refer to the passage:
• 'Healthy eating is a critical component of healthy aging.'
• 'When should one start eating healthy?'
• 'It turns out that it is never too early.'
• From the above-mentioned statements, it is evident that the correct logical inference is that 'eating healthy can be started at any age, earlier the better'.
• Hence, option 2 is the correct answer.

Additional Information

• Critical means to have decisive or crucial importance in the success, failure, or existence of something.
• For eg.- Temperature is a critical factor in successful fruit storage.

Explanation-

Given that,

 P invested Rs. 5000 per month for 6 months of a year.

Q invested Rs. x per month for 8 months of the year.

Total investment by P = 5000 × 6 = 30000 per year

Total investment by Q = 8x per year

Let total profit = P

 Q receives \(\frac{4}{9}\) of the total profit.

So the profit of Q = \(\frac{4P}{9}\)

Profit of P = \(\frac{5P}{9}\)

The profit is shared in proportion to the total investment made in that year.

So, \(\frac{{\frac{{5P}}{9}}}{{\frac{{4P}}{9}}} = \frac{{30000}}{{8x}}\)

\(\frac{5}{4} = \frac{{30000}}{{8x}}\)

x = 3000

Explanation-

Mean- The arithmetic mean of a given data is the sum of all observations divided by the number of observations.

Median - The value of the middlemost observation, obtained after arranging the data in ascending or descending order, is called the median of the data.

Mode - The value which appears most often in the given data i.e. the observation with the highest frequency is called a mode of data.

Given data and Calculation-

Representing the table in tabular form-

 

Mean = \(\frac{{(3 \times 3) + (4 \times 9) + (5 \times 11) + (6 \times 7) + (7 \times 14) + (8 \times 2) + (9 \times 4)}}{{3 + 9 + 11 + 7 + 14 + 2 + 4}}\) = \(292\over50\) = 5.84

Median = \(\frac{{{{\left( {\frac{n}{2}} \right)}^{th}} + {{\left( {\frac{n}{2} + 1} \right)}^{th}}}}{2}\)

\(\frac{{{{\left( {\frac{{50}}{2}} \right)}^{th}} + {{\left( {\frac{{50}}{2} + 1} \right)}^{th}}}}{2} = \frac{{6 + 6}}{2} = 6 \)

Mode = Mark with higher frequency = 7 

So mode > median > mean.

Explanation-

Given that,

The movement will be from P2 to P5.

At the end of each step, it will not occupy any shaded shape.

So the possible shortest path is given below-

P2 to Q4

Q4 to S3

S3 to T5

T5 to R4

R4 to P5.

So the number of steps required is 5.

Explanation:

The Dark Shaded edge is perpendicular to a given line and the Light shaded edge is parallel to the given line so it can be assumed that option 4 is correct but there is no sign of + represented anywhere in the Question.

Hence, the correct answer is "Option 4".

Explanation-

Given equation is, z = sin (y + it) + cos (y - it)

Partially differentiating the equation with respect to y, we have 

\(\frac{{\delta z}}{{\delta y}} = \)cos (y + it) - sin (y - it)

Partially differentiating the equation with respect to y twice, we have

\(\frac{{{\partial ^2}z}}{{\partial {y^2}}} = \) -sin (y + it) - cos (y - it)---------(i)

Partially differentiating the equation with respect to t, we have 

\(\frac{{\delta z}}{{\delta t}} = \) i cos (y + it) + i sin (y - it)

Partially differentiating the equation with respect to y twice, we have

\(\frac{{{\partial ^2}z}}{{\partial {t^2}}} = \) i² sin (y + it) - i² cos (y - it)-------(ii)

Value of i² = -1

So equation (ii) will be, 

\(\frac{{{\partial ^2}z}}{{\partial {t^2}}} = \) cos (y - it) + sin (y + it)------(iii)

\(\frac{\partial^2 z}{\partial t^2}+\frac{\partial^2 z}{\partial y^2}=\) -sin (y + it) - cos (y - it)+cos (y - it) + sin (y + it) = 0

So the partial differential equation derived from the above expression is \(\frac{\partial^2 z}{\partial t^2}+\frac{\partial^2 z}{\partial y^2}=0\).

Explanation-

Order- Order of a differential equation is the order of the highest derivative (also known as differential coefficient) present in the equation.

Degree-

• The degree of the differential equation is represented by the power of the highest order derivative in the given differential equation.
• The differential equation must be a polynomial equation in derivatives for the degree to be defined.

Given data and Analysis-

The equation is given as, \(\frac{d^3y}{dx^3}+x\left(\frac{dy}{dx}\right)^{3/2}+x^2y=0\)

⇒ \(\frac{d^3y}{dx^3}+x^2y=x\left(\frac{dy}{dx}\right)^{3/2}\)

Squaring on both sides the equation will be,

\({\left( {\frac{{{\partial ^3}y}}{{\partial {x^3}}}} \right)^2} + {x^4}y^2 + 2x^2y{\frac{{{\partial ^3}y}}{{\partial {x^3}}}}= {x^2}{\left( {\frac{{dy}}{{dx}}} \right)^3}\)

So highest order = 3

Degree of the highest order derivative = 2

So it is an ordinary differential equation of order 3 and degree 2.

Concepts:

In thin cylindrical shells, there are two principal stresses acts**:**

1. Longitudinal stress: 

\( σ_L = \frac{pd}{4t}\)

2. Hoop Stress:

\( σ_h = \frac{pd}{2t}\)

 Both stresses are tensile in nature.

From the above, it can be said that

σ_h = 2σ_L

The maximum in-plane shear stress is given as:

\(\tau_{max} = \frac{σ _h - σ_L}{2}\)

Calculation:

Given:

σ_h = 30 MPa

σ_L = 30/2 = 15 MPa

Both are tensile in nature.

The maximum shear stress is given as:

\(\tau_{max} = \frac{30 -15}{2}\)

∴ τ_max = 7.5 MPa

Alternate SolutionWe know, 

The maximum shear stress is given as:

\(\tau_{max} = \frac{σ _h - σ_L}{2}\)If we consider 3 - D then minimum Stress = 0

So

⇒  \(\tau_{max} = \frac{30 -0}{2}\) = 15 MPa

 Option A or B is the correct answer as per the official Answer

Concepts:

Modular Ratio:

Modular Ratio in RCC is defined as the ratio between Modulus of Elasticity of Steel and Modulus of Elasticity of Concrete.  It is donated by m.

M = E_s/E_c

Where E_s is the modulus of elasticity of steel

E_c is the modulus of elasticity of Concrete.

As per IS 456 codal provisions:

1. The modulus of elasticity of steel is 2,00 kN/mm² irrespective of the grade of steel.
2. In general, concrete does not have a definite value of modulus of elasticity as it is not a perfectly elastic material.  However, the short-term modulus of concrete can be  taken as:

     Ec = 5000√f_ck

 Where f_ck is the characteristic strength of concrete.

 3. In the working stress method, the modular ratio is assumed to have a value of 280/3σ_cbc where σ_cbc is the permissible compressive stress (in N/mm2) in concrete due to bending.

Concepts:

The pozzolanic materials are those materials which itself do not possess the cementitious properties but when added in certain proportions in ordinary Portland cement, it reacts with a by-product of hydration of cement i.e. with Calcium Hydroxide (Ca (OH)₂ and leads to the formation of CSH gel which posses the cementitious properties.

The pozzolanic reaction is the chemical reaction that occurs in Portland cement upon the addition of Pozzolana and it occurs in between calcium hydroxide and Silicic acid.

The reaction is:

Ca(OH)₂ + Reactive Silicon dioxide + H₂O → C-S-H

Concepts:

Shear center:

• Shear center is a point on the cross-section where the application of loads does not cause its twisting.
• The shear center position is dependent on the cross-section of the beam.

The following points should be remembered for finding the location of shear center.

1. For symmetrical sections, the shear center and center of gravity are the same and coincides   with each other.
2. For unsymmetrical section, the shear center lies on the axis where it has axis of symmetry.
3. For sections made by joining the thin rectangular sections, then shear center lies on the points of intersection of individual symmetrical axis of this rectangular section. 

In this problem point No. 3 is used to find out the location of shear center.

The horizontal thin rectangular section has axis of symmetry along X-axis as shown in the dotted line and the vertical thin rectangular section has an axis of symmetry Y-axis itself (also shown as dotted line) and their point of intersection lies at (0, 3a).

Concepts:

CH à Clay with High liquid limit > 50 %.

ML à Silt with Low Liquid Limit < 50 %.

SP à Poorly graded sand with fines < 5%.

SW à Well graded sand.

Among the above, Clay and silt are fine-grained soils and Sand is coarse-grained soil.

Generally, coarse-grained soil has higher permeability than fine-grained soil. Among the silt and clay; silt has more permeability than that of Clay. Well-graded soil has better permeability than poorly graded.

Based on these facts, the decreasing order of permeability of soil is:

SW > SP > ML > CH

Concepts:

The soil that is retained at a slope steeper than it can sustain by virtue of its shearing strength, exerts a force on the retaining wall. This force is called earth pressure.

The ratio of the horizontal stress to the vertical stress is called the coefficient of Earth pressure and the same ratio is called the coefficient of active earth pressure and coefficient of passive Earth pressure when soil is in Rankin’s active and passive state respectively.

If the wall is pushed towards the backfill, the soil is compressed and the soil offers its resistance by virtue of its shearing resistance. Since shearing resistance builds up in the direction of the wall the earth pressure gradually increases and reaches a value where backfill cannot withstand and slip surface is formed and force acting on the wall at this stage is called passive earth pressure. Mathematically, it is given as:

K_p = σ_h/σ_v

K_p > 1 always

∴ σ_h > σ_v

Note:

When the wall is moving away from the soil, the soil mass expands, resulting in a decrease in earth pressure due to mobilization of shearing resistance. A portion of the backfill tends to break away from the rest of the soil mass and tends to move downward and outward relative to the wall.  A stage is reached when the entire shearing resistance is mobilized and at this stage, the force acting on the wall is called active earth pressure.

 Mathematically, it is given as:

K_a = σ_h/σ_v

K_a < 1 always

∴ σ_h < σ_v

Concepts:

Mach number is dimensionless quantity  and it is defined as the ratio of flow velocity past a boundary to the local speed of sound. It is related to compressibility of fluid flow.

Reynolds Number is dimensionless quantity  and it is defined as ratio of inertia force to viscous force in fluid flow. Based on this flow can be considered as either laminar or turbulent.

Froude Number is dimensionless quantity  and it is defined as square root of ratio of inertia force to gravitational force. It governed the fluid flow in open channels.

Euler Number is dimensionless quantity  and it is defined as square root of ratio of inertia force to pressure force. It governs the fluid flow in pipes.

Concepts:

The following points should be remembered for the hydraulic conductivity of unsaturated soil.

1. The hydraulic conductivity of the unsaturated soils is the non-linear function of soil suction head (ψ) and soil moisture content (θ).
2. The hydraulic conductivity for coarse-textured soil is higher than for finer-textured soil.
3. On increasing the water content, the soil suction head (ψ) decreases, and hydraulic conductivity (K) increases.

It depends on the soil pore geometry as well as the fluid viscosity and density.

The hydraulic conductivity for a given soil becomes lower when the fluid is more viscous than water

Explanation-

• When two-channel sections have different bed slopes the condition is called a break in grade.
• Under this situation following conditions must be remembered for drawing the flow profile.
• CDL is independent of the bed slope.
• The steeper the slope lesser is the normal depth of flow.
• Flow always starts from NDL and tries to meet NDL.
• Subcritical flow has downstream control and supercritical flow has upstream control.
• Steep slope-

 

Given data and Analysis-

• Flow is changed from steeper to sleep.
• So the Normal depth of flow will increase in a steep slope, as the slope is decreased.
• CDL will remain the same for both the slope.
• So from the figure shown below it can be concluded that flow will be S₃ profile.

Concepts:

The total hardness in milligram equivalent per liter is ratio of hardness in milligram per liter as CaCO₃ to equivalent weight of CaCO₃.

Further, The equivalent weight of CaCO₃ is the ratio of molecular weight to valency factor.

Calculation:

The valency factor for CaCO₃ is 2 as Ca can loose maximum 2 electrons and converted to Ca⁺² and CO₃^2- ions when dissolved in water.

The molecular weight of CaCO₃ is = 40 + 12 + 3 × 16 = 100 gm

The equivalent weight of CaCO₃­ is = 100 /2 = 50 gm

∴ The total hardness in milligram equivalent per liter  = 500/50 = 10.

Concept:

• The scale of a photograph is determined by the focal length of the camera and the flying height above the ground. The focal length is the distance from the middle of the camera lens to the focal plane.
• The scale of a photo is equal to the ratio between the camera's focal length and the plane's altitude above the ground level (AGL) being photographed. If the focal length and flying altitude above the surface is known, the scale can be calculated using the following formula:

​

 

Scale = \(\rm \frac{f}{H}\)

f = Focal length of the camera

H = Flying height above ground level

 

Flying Height Above Ground (AGL) = Altitude above sea level (MSL) - Average elevation of the terrain

Calculation:

Flying height = 3.5 km = 3500 m above mean sea level

focal length of camera = 152 mm

Average ground elevation = 460 mm above mean sea level

Scale = \(\frac{152}{(3500-460)1000}\)= 1 ∶ 20000

Concept:

Correction for measurement with faulty tape-

Let l be the true length of the tape and l' be the faulty length of the tape, then 

The true length of the tape x True length of tape = Faulty length of tape x Faulty length of tape

Slope Correction-

• When a measurement is taken along an inclined plane( along the natural slope of the ground), the measured distance is greater than the horizontal distance.

• The difference between the slope distance and the horizontal distance is called slope correction.

• This correction is always negative.

• Where h= difference in elevation between the ends

• L= Inclined length measure, l=Horizontal length

• C_s= Correction due to sag.

• Correction due to slope = L(cos θ - 1)

Calculation:

Slope of line = 1 in 5

tan θ =\(\frac{1}{5}\)

θ =11.31 degree

Correction due to slope = 350 ( cos 11.31 - 1)=-6.8 m

Nominal length of tape = 50 m

Tape is short by 0.1 m, so faulty length of tape = 49.9 m

So true length of line = \(\frac{(350)(49.9)}{50}\)=349.3 m

Correction = 349.3-350 = -.7 m

Net correction = -(6.8+.7)=-7.5 m

So corrected length of measured line = 342.5 m

Concept:

• Eigenvector of a matrix A is a vector represented by a matrix X such that when X is multiplied with matrix A, then the direction of the resultant matrix remains same as vector X.
• Mathematically, the above statement can be represented as AX = λX

Properties of eigenvalues-

Eigenvalues of unitary and orthogonal matrices are of unit modulus |λ| = 1

• If λ1, λ2…….λn are the eigenvalues of A, then kλ1, kλ2…….kλn are eigenvalues of kA
• If λ1, λ2…….λn are the eigenvalues of A, then 1/λ1, 1/λ2…….1/λn are eigenvalues of A-1
• If λ1, λ2…….λn are the eigenvalues of A, then λ1k, λ2k…….λnk are eigenvalues of Ak

Eigenvalues of A = Eigen Values of AT (Transpose)

Sum of Eigen Values = Trace of A (Sum of diagonal elements of A)

Product of Eigen Values = |A|

Maximum number of distinct eigenvalues of A = Size of A

Calculation:

The given eigenvalues are 5 and −2.

Q = M3 - 4M2 - 2M

Eigen values of M³ are 125 and -8,4M²  are 100 and 16, 2M are 10 and -4.

SO eigen values of Q, λ1= 125-100-10 = 15,λ2 = -8-16-(-4)=-20

So the eigen values are 15 and -20.

Concept:

BOD, COD, TOC, and TOD all are associated with the organic content of water. COD is related to both the inorganic and organic content of water.

 

Chemical Oxygen Demand is always more than Biological Oxygen demand as COD includes both bio-degradable and non-biodegradable organic matter.

If the non-biodegradable matter is absent then COD may be equal to BOD.

But ultimate BOD is more than BOD5.

So COD ≠ BOD5 and COD > BOD5.

Important Points

• A high COD/BOD ratio means the amount of COD is more as compared to the amount of BOD.
• Similarly, High BOD/COD ratio means the amount of BOD is more as compared to the amount of COD.

Concept:

Microfiltration and Ultrafiltration

• Microfiltration is the process of physically removing suspended solids from water, through a membrane.
• Ultrafiltration blocks everything microfiltration can with the addition of viruses, requiring a slightly higher pressure to achieve this.
• UF and MF can be used in the following processes:
• Treating wastewater
• Concentrating proteins
• Chemical process separation
• Separating oil/water emulsions
• Removing pathogens from milk

Electrodialysis

• Electrodialysis (ED) is a membrane technique, during which ions are transported through a semipermeable membrane, under the influence of an electric potential. It is a very versatile technology for the separation of difficult mixtures.
• ED is very useful for water treatment, aiding in the removal of mineral salts, sulfate, nitrate, etc. from brackish water and seawater. ED is also useful for wastewater reduction or recovery.

Reverse Osmosis

• Reverse osmosis is the process in which pressure is applied to overcome colligative property and osmotic pressure that is directed by a thermodynamic parameter and a chemical difference of a solvent.
• This application is mainly applied in the production of potable water in water plants and in industries.
• It is used for the desalination of brackish water and seawater and it is also used for the treatment of wastewater.

Explanation:

In a horizontal curve, e+f = V²/Rg 

 R = V²/g (e + f)

Where

R = Radius of curvature, V = design speed

e = super elevation, f = Coefficient of friction, g = Acceleration due to gravity

Given data and analysis:

It is required to decrease the radius of curvature.

As,  R = V2/g (e + f)

From the above formula, R is directly proportional to velocity and inversely proportional to super-elevation.

So to decrease the radius of curvature decrease the design speed and increase the super-elevation

Concept:

Given iteration scheme is \(x_{n+1}=\frac{1}{2}\left(x_n+\frac{P}{x_n}\right)\)

The iteration scheme will converge when x_n+1 = x_n

\(\rm x_{n +1} = \frac{1}{2} \left(x_n + \frac{P}{x_n} \right)\)

Converges when x_n+1 = x_n = α

\(\rm α = \frac{1}{2} \left( α + \frac{P}{α} \right)\)

\(\frac{α}{2} = \frac{P}{2 α}\)

α² = P

α = √P

Given recursive relation is finding square root of P

When P = 2, x₅ = √2 = 1.4124 

When P = 3, x₅ = √3 = 1.732 

Similarly when P = 10,

x₅ = √10 = 3.162

Ans = 3.162

Concept:

cos 2x = 2 cos²x - 1

cos²x = \(\rm \left( \frac{1 + \cos 2x}{2} \right)\)

Calculation:

Given that,

\(\rm f(x) = \sum_{n = 0}^\infty f_n \cos nx\)

f(x) = cos⁴ x

⇒ cos⁴x = cos²x . cos²x

\(\rm = \left( \frac{1+ \cos 2x}{2} \right). \left( \frac{1+ \cos 2x}{2} \right)\)

\(\rm = \frac{1}{4} [ 1 + 2 \cos 2x + \cos^2 2x]\)

\(\rm = \frac{1}{4} \left[ 1 + 2 \cos 2x + \frac{(1+ \cos 4x)}{2} \right]\)

\(=\rm \frac{1}{4} + \frac{\cos 2x}{2} + \frac{1}{8} + \frac{\cos 4x}{8}\)

\(= \rm \frac{3}{8} + \frac{\cos 2x}{2} + \frac{\cos 4x}{8}\)

when n = 4, F₄ = 1/8, F₅ = 0 [As no term as cos 5x]

so, F₄ + F₅ = 1/8 = 0.125

Concept:

The volume of soil solids remains constant but the total volume of soil may vary according to different situations.

So (V_s)_heap = (V_s)_{compacted soil}

V_s = Volume of soil solids

V_s = \(\frac{Total-vol-of-soil}{1+e}\)

Calculation:

An un-compacted heap of soil has a volume = 10000 m3 and void ratio = 1

The compacted volume of soil = 7500  m3

So,

(V_S)_heap = (V_S)_{compacted soil}

So, \(\rm \left( \frac{V_T}{1 + e}\right)_{heap} = \left( \frac{V_T}{1 + e}\right)_{compacted \ soil} \)

\(\frac{10000}{1+1} = \frac{7500}{1+e}\)

e = 0.5

Concept:

Boussinesq solved the problem of stress distribution in soils due to a concentrated load acting at the ground surface, by assuming a suitable stress function. The following assumptions are made in the solutions by the theory of elasticity,

• The soil mass is an elastic medium, for which the modulus of elasticity is constant.
• The soil mass is homogenous, isotropic, semi-infinite.

Stress due to a point load  at any depth at point z is given by,

\(\rm \sigma_Z= K_B \frac{Q}{z^2}\)

K_B = Boussinesq influence factor

\(\rm = \frac{3}{2 \pi} \left( \frac{1}{1 + \left( \frac{r}{z} \right)^2}\right)^{5/2} \)

Where r = radial distance of the point as shown in figure, z= depth at which stress is required

Calculation:

The magnitude of point load = 3000 kN

P and Q are at depths 1 m and 2 m below the ground along the line of application of load.

So r = 0 for both the cases.

\(\rm (\sigma_z)_p = \frac{3}{2 \pi} \frac{Q}{(1)^2}\)

\(\rm (\sigma_z)_p = \frac{3}{2 \pi} \frac{Q}{(2)^2}\)

\(\rm\frac{ (\sigma_z)_p}{(\sigma_z)_q }= \frac{\frac{3}{2 \pi} \frac{Q}{(1)^2}}{\frac{3}{2 \pi} \frac{Q}{(2)^2}}=4\)

Explanation-

Friction ratio is defined as the ratio between side friction and point resistance.

Friction ratio = \(\frac{{{\rm{Side \ friction}}}}{{{\rm{Tip \ resistance}}}}\)

Given data and Calculation-

Friction ratio = 1%

point (tip) resistance qc = 1000 kPa

sleeve (side) friction  = 0.01 × 1000 = 10 kPa

The Data is insufficient, Marks are given to all.

Explanation-

In question, the root zone depth of the crop is given, but there is no information about the equivalent depth of available water. So, the frequency of irrigation can’t be determined.

Consumptive use-

• Consumptive use of water by a crop is the depth of water consumed by the plant in the process of interception, transpiration, and evaporation during crop growth.
• These two processes are collectively called evapotranspiration.
• Consumptive use of water by a crop is expressed as the depth of water per unit area for a specified period, such as days, months, or seasons.

Water conveyance efficiency-

• It is the ratio of the amount of water applied, to the land to the amount of water supplied from the reservoir.

Water Application efficiency

• It is the ratio of the water stored in the root zone of plants to the water applied to the land. It is obtained by the expression.

Explanation-

Surveyors express angles in several ways. When specifying directions, as is done in the preparation of a property survey, angles may be specified as bearings or azimuths.

A bearing is an angle less than 90° within a quadrant defined by the cardinal directions.

Azimuth is an angle between 0° and 360° measured clockwise from the North.

"South 45° East" and "135°" are the same direction expressed as a bearing and as an azimuth.

 

Given data and Explanation-

The bearing of a survey line =  N31°17'W.

From the figure, bearing = 360°00' - 31°17' = 328°43' = 328.72

Concept:

In 3D, rotations can also be defined as linear transformations, a rotation in 3D can be represented by a matrix equation P'=RP where R is a rotation matrix.

R =  \(\begin{bmatrix}1&0&0\\ 0 &\cos \theta&-\sin \theta\\ 0&\sin \theta&\cos \theta\end{bmatrix}\)

where θ is the angle of rotation with the positive axis in the clockwise direction.

Given Data and Calculation:

Given that, Coordinate of point P = (1,1,1)

The angle of rotation = 45°

New Coordinate will be,

\(\rm\begin{bmatrix}x'\\ y'\\ z'\end{bmatrix}=\begin{bmatrix}1&0&0\\ 0 &\cos \theta&-\sin \theta\\ 0&\sin \theta&\cos \theta\end{bmatrix}\rm\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}\)

= \(\begin{bmatrix}1&0&0\\ 0 &\frac{1}{\sqrt2}&-\frac{1}{\sqrt2}\\ 0&\frac{1}{\sqrt2}&\frac{1}{\sqrt2}\end{bmatrix}\rm\begin{bmatrix}1\\ 1\\ 1\end{bmatrix}=\begin{bmatrix}1\\ 0\\ \sqrt2\end{bmatrix}\)

So the solution will be 1.rotation matrix.p′=Rp

Concept-

Shear force-

It is defined as the algebraic sum of all the vertical forces, either to the left or to the right hand side of the section. 

Bending moment -

It is defined as the algebraic sum of the moments of all the forces either to the left or to the right of a section

Axial force-

 Axial force can be defined as the force acting on a body in its axial direction.

Given data and calculation:

Let the axial force and shear force acting at section Q is represented by A and V respectively.

Bending at section Q = Force x perpendicular distance = \(\rm \frac{PR}{√ 2}kNm,\)

Net vertical force acting at section Q = 0

Net horizontal force acting at section Q = P

So FBD of the section Q will be as shown in figure below-

By taking components of forces in horizontal and vertical directions the equations obtained are -

A cos 45° - V sin 45° = P

\(\rm \frac{A}{√2}-\frac{V}{√2}=P\)      ......(1)

A sin 45° + V cos 45° = 0

\(\rm \frac{A}{√2}+\frac{V}{√2}=0\)       ........(2)

⇒ \(\rm \frac{A}{√2}=-\frac{V}{√2}\)

⇒ A = -V     .........(3)

\(\rm \frac{A}{√2}+\frac{A}{√2}=P\)

\(\rm \sqrt2A=P\)

⇒ \(\rm A=\frac{P}{\sqrt2}=-V\)

So the magnitude of axial force and shear force are \(\rm \frac{P}{√ 2}kN, \frac{p}{√ 2}kN\).

Important Points As the beam is curved and a single horizontal force is acting and the section is not at 90 degrees, bending moment, shear force, and axial force will exist. None of the above three will be zero.

• The only option that exists which has all the three values as non-zero is option 1. To save time without any calculation tick option 1.

Concept-

Strength of weld = Permissible stress in weld × Effective area of the weld.

Strength of any section = Permissible stress in section × Area of the cross-section

The effective area of the weld = Throat thickness × Effective length

For the angle section, the section is not symmetrical, so the length of the weld will be more along the length which is nearer to the centroid as shown in the figure.

The ratio of the Length L1 and L2 will be given as,

\(\frac{L1}{L2}\)=\(\frac{B-Cyy}{Cyy}\)

Where B= length of the Connecting leg

C_yy = Distance of the centroid from the edge of the angle section

Given Data and Calculation-

Permissible shear stress on the section through the throat of the fillet weld = 108 MPa

permissible stress in the angle = 150 MPa 

Cyy = 28.4 mm

Area of cross-section of the angle = 1903 mm2 

strength of the section = 150 × 1903=285.45 kN

effective size of the weld = 0.7 × 5 = 3.5 mm

Area of the weld required = \(\frac{285450}{108}\)=2643.055 mm²

Length of the weld required = \(\frac{2643.055}{3.5}\)=755.158 mm

\(\frac{L1}{L2}\)=\(\frac{71.6}{28.4}\)=2.52

L₁ = 2.52L₂

3.52L₂ = 755.158 mm

L2 = 214.53 mm

L1 = 541 mm

So L₁ and L₂ are 541 and 214 mm respectively.

 Important Points

• After Obtaining the length of the weld required check the sum of the weld lengths given in options.
• Only option 1 and 2 are matching with the required length of the weld, and it is known that L1 will be more than L2. So tick option 1 with out doing further calculation.

Concept-

The critical path method (CPM) is a resource-utilization algorithm for scheduling a set of project activities.

The essential technique for using CPM is to construct a network diagram of the project that includes the following-

• A list of all activities required to be completed
• The dependencies between the activities.
• The estimate of the time duration that each activity will take to complete.

Rules for constructing a network diagram-

• The direction of the arrows should flow from left to right avoiding mixing of arrows.
• No single activity can be represented more than once in a network diagram.
• Before an activity can be undertaken, all activities preceding it must be completed. The logical relationship has to be maintained.
• Dummy activity must be used when required.
• A network diagram must have one start and one end event.

 

Given data and Calculation-

 

As per the relationships given in the question, the network diagram will be-

So critical path is Q to S and the longest duration will be 22 days.

​

Concept:

Normally consolidated soil-

• Normally consolidated clays are these that are currently experiencing the maximum vertical overburden effective pressure they have ever experienced in their history.
• Over consolidated clays have experienced higher overburden stress in the past.
• OCR = (maximum overburden effective stress)/ (Currrent overburden effective stress)
• So for Normally consolidated soil OCR = 1.

 

Sensitivity of clay-

• The ratio of unconfined compression strength of clay in the natural or undisturbed state to that in the remolded state, without any change in the water content is called as sensitivity of soil.
• S_t = S_u (Undisturbed)/S_u (remoulded)

• So for Quick clay sensitivity is more than 16.

Sand in Critical State- 

• The dilation angle of sand at the critical state is zero.

Type of clay according to plasticity-

• According to IS classification of soil clay with a liquid limit of more than 50 is classified as high plasticity, between 35 to 50 is classified as medium plasticity and less than 35 is classified as low plasticity**.**

So the correct option is 1.

Explanation:

According to Rankine's Theory of Earth Pressure, the inclination of the failure plane with respect to the minor principal stress direction is given by:

\( (45^\circ + \frac{\phi}{2}) \)

where:

• ϕϕ

  = Angle of internal friction of the soil.

Application in Active and Passive States:

The above equation holds true for both active and passive earth pressure states because:

• In Active Earth Pressure State:
• The major principal stress acts vertically.
• The minor principal stress (lateral earth pressure) acts horizontally.
• The failure plane is inclined at \( (45^\circ + \frac{\phi}{2}) \) to the minor principal stress direction.
• In Passive Earth Pressure State:
• The major principal stress acts horizontally.
• The minor principal stress (lateral earth pressure) acts vertically.
• Again, the failure plane follows the same inclination of \( (45^\circ + \frac{\phi}{2}) \).

Since this failure plane inclination applies to both active and passive earth pressure conditions, the correct answer is:

Concept-

Henry’s law-

Henry’s law is a gas law that states that at the amount of gas that is dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid when the temperature is kept constant.

The constant of proportionality for this relationship is called Henry’s law constant (usually denoted by ‘kH‘). The mathematical formula of Henry’s law is given by: C ∝ P (or) C = kH.P

Where, 

‘P’ denotes the partial pressure of the gas in the atmosphere above the liquid. ‘C’ denotes the concentration of the dissolved gas. ‘kH’ is Henry’s law constant of the gas.

If the constant is defined in terms of solubility/pressure, it is referred to as Henry’s law solubility constant (denoted by ‘H’).

On the other hand, if the proportionality constant is defined in terms of pressure/solubility, it is called Henry’s law volatility constant (denoted by ‘kH’).

Given data and Calculation-

Henry's law constant (kH) = 1.3 \(\frac{mmol}{lit-atm}\)​

Partial pressure of O2 in the atmosphere = 0.21 atm

Molecular weight of O2 = 32 g/mol

O₂ dissolved = 1.3 × 0.21 = 0.273 mmol/litre

Molar concentration = \(\frac{given- mass}{Molecular- mass}\)

Mass of O₂ = 0.273 × 10^-3 × 32 = 8.736 mg/lit​

Concept-

• Hardness is the concentration of multivalent cations present in the water. It may be induced due to Ca²⁺, Mg²⁺, Fe^3+, etc.
• Hardness due to carbonate and bicarbonate of multivalent cations is called carbonate hardness. This hardness is known as temporary hardness as it can be removed by simple boiling or the addition of lime.
• Total hardness = Carbonate hardness + Noncarbonate Hardness
• Total hardness as CaCO₃ (in mg/L) = \(\frac{{\left[ {C{a^{2 + }}} \right] × {\rm{equivalent weight of CaC}}{{\rm{O}}_3}}}{{{\rm{equivalent weight of C}}{{\rm{a}}^{2 + }}}} + \frac{{\left[ {M{g^{2 + }}} \right] × {\rm{equivalent weight of CaC}}{{\rm{O}}_3}}}{{{\rm{equivalent weight of M}}{{\rm{g}}^{2 + }}}}\)
• Carbonate hardness = minimum { Total hardness, Alkalinity}

Given data and Calculation-

Concentrations of Ca²⁺ = 100 mg/L,eq wt = 20

Concentrations of Mg^₂₊ = 36 mg/L, eq wt = 12

Concentrations of HCO^-₃ = 122 mg/L

Alkalinity = 122/61 = 2 × 50 = 100 mg/L as CaCO₃ 

Total hardness as CaCO3 (in mg/L) = \(\frac{{100 \times 50}}{{20}} + \frac{{36 \times 50}}{{12}}\)=400 mg/L as CaCO₃

Carbonate hardness = minimum { 400, 100} = 100 mg/L as CaCO₃

Concept-

• The fundamental relation between flow(q), density(k) and mean speed v is q = k x v

• When the density is zero, flow will also be zero, since there is no vehicles on the road.

• When the number of vehicles gradually increases the density as well as flow increases.

• When more and more vehicles are added, it reaches a situation where vehicles can't move. This is referred to as the jam density or the maximum density. At jam density,  flow will be zero because the vehicles are not moving.

• There will be some density between zero density and jam density, when the flow is maximum. The relationship is normally represented by a parabolic curve as shown in figure.

 

Given data and Analysis -

 

• At point S, speed is non zero but flow is zero, so density will be zero to get zero flow. So K_s will be minimum.
• At point P, speed is zero which means it is the jam condition so the density will be maximum.SO Kp will be maximum.
• So from point S to point P the density will increase gradually.
• So the correct order of density will be  \(k_P>k_Q>k_R>k_s\).

Explanation-

Continuity of a function-

• A function f(x) is said to be continuous at x=a if \(\mathop {\lim }\limits_{x \to a} f(x) = f(a)\).
• A function is said to be continuous on the interval [a,b] if it is continuous at each point in the interval.

Local maxima and minima-

• Local maxima and minima are the maxima and minima of the function which arise in a particular interval.
• Local maxima would be the value of a function at a point in a particular interval for which the values of the function near that point are always less than the value of the function at that point.
• Whereas local minima would be the value of the function at a point where the values of the function near that point are greater than the value of the function at that point.

 

Differentiability of a function-

• If the Left-hand derivative and the Right-hand derivative at a point are equal then the function is said to be differentiable at that point.
• When there is no sharp edge in the graph of the function the function is differentiable in that domain.

 

Given data and Calculation-

The given function is f(x) = max{3 - x, x - 1}

The graph of the function will be,

From the above graph, it is clear that it is continuous throughout its domain.

It has a local minimum at x = 2.

As there is a sharp edge at point (2,1) so it is not differentiable at x = 2.

So option 1 and 2 are correct.

Concept-

• Frictional Force refers to the force generated by two surfaces that contact and slide against each other.
• The full amount of friction force that a surface can apply upon an object can be easily measured with the use of the given formula:

F_frict = μ F_{norm ,} Where μ  = coefficient of friction and F_norm = Normal reaction

 

 

 

• When the push or pull force is more than the full amount of friction force then sliding will occur.
• When the force does not act the centre of gravity of a body, then the body will undergoes rotation which may result in overturn of a body.

Given data and Analysis-

 

[FBD of body]

Applying equilibrium equation in vertical direction, N = W = 25 kN

Minimum force for sliding = P_min = (f_s)_max

(f_s)_max = μ N = 0.3 × 25 = 7.5 kN

 

As the force is not acting on C.G of the body, there is a possibility of overturning.

So taking moment about A,

25 × 0.5 = P_min ×  2

⇒ P_min = 6.25 kN

 

• Force required for overturning is less than the force required for sliding, so first overturning will take place and sliding will never occur.
• No motion occurs for P ≤ 6 kN, as the minimum force required for overturning is 6.25 KN.
• So option 1, 2 and 3 are correct.

Concept:

For conveying an irrigation canal across a natural channel is by providing water conveying structure which may:

• Carry the canal over the natural stream;
• Carry the canal beneath the natural stream; or
• Carry the canal at the same level of the natural stream.

Type 1- Cross drainage work carrying the canal over the drain-

• Aqueduct-
• In an aqueduct, the canal bed level is above the drainage bed level so the canal is to be constructed above the drainage.
• Natural drain water goes under the irrigation canal.
• Syphon Aqueduct-
• In a syphon aqueduct, canal water is carrier above the drainage but the high flood level (HFL) of drainage is above the canal trough. The drainage water flows under syphonic action and there is no presence of atmospheric pressure in the natural drain.

Type 2- Carry the canal beneath the natural stream-

• Super passage-
• Super passage structure carries drainage above canal as the canal bed level is below drainage bed level.
• Natural drain water goes over the irrigation canal.
• Canal Syphon-
• In a canal syphon, drainage is carried over a canal similar to a su  per passage but the full supply level of the canal is above than the drainage trough. so the canal water flows under syphonic action and there is no presence of atmospheric pressure in the canal.

​Type 3- Carry the canal at the same level of the natural stream.

• Level Crossing-
• When the bed level of the canal is equal to the drainage bed level, the level crossing is to be constructed.
• Natural drain water goes through the irrigation canal.

Concept-

• Euler's method is an approximate method that is used to find out the solution of the first-order differential equation.
• Let any first-order differential equation is represented as
• \(\rm \frac{dy}{dx}=f(x,y)\)
• Initial values of x and y are represented as X₀ and Y₀.
• So as per Euler's formula-
• y_n+1 = y_n + n f(x_n, y_n)

n = step size

Given data and Calculation-

\(\rm \frac{dy}{dx}=4(x+2)-y\)

f(x, y) = 4(x + 2) - y

y₀ = 3 & x₀ = 1.

n = 0.2

y₁ = y₀ + n f(x₀, y₀)

 = 3 + 0.2 × f(1,3)

= 3 + 0.2 [4(1 + 2) - 3]

= 3 + 0.2 [9] = 4.8

y₂ = y₁ + n f(x₁, y₁)

= 4.8 + 0.2 × f(1.2, 4.8)

= 4.8 + 0.2 [4(1.2 + 2) - 4.8] = 6.40

Concept:

The normal equation for Fitting a straight line by the least square method is:

Σy = na + b Σx

Σxy = a Σx + b Σx²

Where

n = Total number of observations, a and b are the coefficients.

By solving the above two equations coefficients a and b can be obtained.

Given Data and Calculation:

n = 4 So

51 = 4a + 14b

188 = 14a + 54b

Solving the above two equations a = 6.1 and b = 1.9

Concept-

Method of sections-

• The method of sections is a process used to solve for the unknown forces acting on members of a truss.
• The method involves breaking the truss down into individual sections and analyzing each section as a separate rigid body.
• Steps to use the method of sections-
• Treating the entire truss structure as a rigid body, draw a free body diagram, write out the equilibrium equations, and solve for the external reacting forces acting on the truss structure.
• Cut the truss into two separate sections. The cut should include the member in which force is required to be calculated. The cut does not need to be a straight line.
• Draw a free body diagram for either one or both sections that is created after cutting.
• Solve the equilibrium equations and get the unknown values.

 

Given data and calculations-

Taking moment about J,

-H_A × 1 + 8V_A - 70 × 4 = 0

8V_A - H_A = 280   ....(1)

By cutting the section at (1) - (1)

∑M_E = 0

⇒ V_A × 4 - H_A × 4 = 0

⇒ V_A = H_A   ....(ii)

From equation (i) & (ii)

7H_A = 280

⇒ H_A = 40 kN ⇒ H_j = 40 kN.

V_j = 70 - 40 = 30 kN

It is required to find the force in member EF,

considering right hand side,

F_EF = V_j ⇒ F_EF = 30 kN

Concept:

Stiffness-

Stiffness for a member at a joint is the moment or force required to produce unit rotation or displacement at that joint.

• Stiffness of a member if the far end is fixed = 4EI/L
• Stiffness of a member if the far end is hinged = 3EI/L
• Stiffness of a member if the far end is free = 0
• Stiffness of a member if the far end is guided roller = EI/L

Where EI = flexural rigidity of the beam section

Distribution factor-

• The factor by which the applied moment is multiplied to obtain the end moment at any member is known as the distribution factor.
• Distribution factor = \(\frac{member stiffness}{Joint stiffness}\)

 

Carry over moments-

• The distributed moments in the ends of members meeting at a joint cause moments in the other ends, which are assumed to be fixed. These induced moments at the other ends are called carry-over moments.

When the far end is a fixed carry over moment will be M/2,

When the far end is guided roller carry-over moment will be M.

When the far end is roller carry over moment will be 0. Where Applied moment = M.

Given data and calculation:

For the beams as shown in the figure,

 

Length of FK and FG and FH = 5 m

flexural rigidity EI = 105 kN-m2

Stiffness of FK and FG and FH = 4EI/L

Distribution factor for member FH = 1/3

So moment carried by FH at joint F = M/3

Carry over moment at joint H = M/6

It is given that Moment at Joint H = 10 kNm.

so M/6 =10

M = 60 kNm.

Concept:

Influence line diagram-

• An influence line diagram of a structural response (axial force, bending moment, Shear force) is the line representing the effect of a unit loading action(Unit point load, moment etc.) applied at every possible location in the structure.
• It can be used to find a maximum bending moment, shear force, etc.

Influence line diagram for a simply supported beam-

 

 

• The maximum bending moment point will be under the unit load.
• The bending moment will be obtained by multiplying the magnitude of load with the ordinate of ILD.

 

Given data and Calculation:

 

The loading condition is as shown in figure.

ILD for bending moment R is shown below-

Ordinate of ILD at point R = 4/5 = 0.8

Ordinate of ILD at a point which is at a distance of 3m from R = 1/5 = 0.2

So maximum bending moment = (200 × 0.8) + (0.2 × 100) = 180 kNm.

Concept:

The (average) design shear strength τ_c of concrete in reinforced concrete beams without shear reinforcement is limited to the value of the nominal shear stress τ_v corresponding to the load at which the first inclined crack develops.

The magnitude of design shear strength depends upon the grade of concrete and the percentage of tensile reinforcement.

For prismatic members of rectangular (or flanged) sections, the IS 456-2000 (Cl. 40.1) uses the term nominal shear stress

, defined at the ultimate limit state, as follows: 

\(\rm τ_ν=\frac{V_u}{bd}\)

where V_u =  factored shear force at the section under consideration,

b = width of the beam (taken as the web width, b_w in flanged beams), and

d = the effective depth of the section.

If V_uc and V_us denote respectively the ultimate shear resistance of the concrete and the shear reinforcement, then the total ultimate shear resistance V_uR at any section of the beams is given by V_uR = V_uc + V_us and V_uc = τ_cbd

From a design viewpoint, suitable shear reinforcement has to be designed if the factored shear V_u exceeds V_uc (i.e., τν exceeds τc), and the shear resistance required from the web reinforcement is given by .V_us ≥ V_u - V_uc = (τ_{ν -} τ_c) bd

for vertical stirrups, the shear resistance V_us is obtained as V_us = 0.87 f_yA_sν d/S_ν

 Where A_sv = the total cross-sectional area of one stirrup, fy = Yield strength of the tension reinforcement,

d = effective depth of the beam, S_v = Spacing of stirrups

Given data and Calculation:

b = 300 mmd = 580 mm

Grade of concrete = M30, grade of steel = 415

Percentage of tension reinforcement = 1%

Design shear strength of beam = 0.66 N/mm²

Factored shear force = 440 kN

Stirrups specification = 2-legged 10 mm diameter vertical stirrups,

\(\rm A_{sv}=2× \frac{\pi}{4}×10^2\) = 157.079 mm²

\(\rm τ_v=\frac{V_u}{bd}=\frac{440×10^3}{300×580}\) = 2.528 N/mm²

τ_c = 0.66 N/mm²

τ_v > τ_c

So, V_uc = τ_c bd = 0.66 × 300 × 580 = 114.480 kN

Shear force nesisted by shirrups = (τ_v - τ_c) bd

= (2.528 - 0.66) × 300 × 580

= 325.032 kN

For spacing of shirrups,

325.032 × 10³ = \(\rm \frac{0.87\times415\times157.079\times580}{S_v}\)

S_v = 101.2 mm

Concept-

As per IS 2911 (Part I)-1979, the ultimate load capacity of a pile in cohesive soil (clay) is given by –

Q_u = C N_c A_b + α c̅ A_s 

where,

C = Cohesion at the pile tip (kPa);

N_c = Bearing capacity factor, (equal to 9 for piles) , Ap = the cross-sectional area of pile toe ;

α = reduction factor also called shear mobilization factor or adhesion factor;

c̅ = average cohesion over the pile length; and ,

As = surface area of pile shaft

For clay, cohesion = \(\frac{Unconfined\ Compressive\ strength}{2}\)

Calculation:

Given:

Length of pile = 10 m

Unconfined compressive strength of clay = 100 kPa

Cohesion = \(\frac{100}{2}\) = 50 kPa

Bearing capacity factor N_c = 9;

Adhesion factor α = 0.7

Ultimate load capacity of pile = 632 kN

Let the width of pile = B

Surface area of pile = 4BL = 40B

Cross sectional area of pile tip = B²

632 = 50 × 9 × B² + 0.7 × 50 × 4B × 10

→ 632 = 450 B² + 1400 B

→ 450 B2 + 1400B - 632 = 0

→ B = 400 mm

Concept:

The time factor T_v is constant for a given degree of consolidation and given boundary conditions of t he problem under consideration, the time required to attain a certain degree of consolidation is directly proportional to the square of its drainage path and inversely proportional to the coefficient of consolidation.

Time required for consolidation,  \(\rm t=\frac{d^2}{C_v}T_v\)

where d= length of drainage path, C_v = Coefficient of consolidation, T_v = time factor 

When U < 60%: \(\rm T_v=\frac{\pi}{4}\left(\frac{U%}{100}\right)^2\)

When U > 60%: \(\rm T_v=-0.9332 \log_{10}\left(1-\frac{U%}{100}\right)-0.0851\)

or T_v = 1.7813 - 0.9332 log₁₀ (100 - U%)

Where U = Degree of consolidation. (in percentage)

Given data and Calculation-

Given that, 

(C_v)_field = (Cv)lab  As the soil is the same.

Concept:

Unit hydrograph-

• A unit hydrograph is defined as the hydrograph of direct runoff resulting from one unit depth (1 cm) of rainfall excess occurring uniformly over the basin and at a unit form rate for a specified duration ( D hours).
• The term unit here refers to a unit depth of rainfall excess which is usually taken as 1 cm.

S-curve-

• If it is desired to develop a unit hydrograph of duration mD, where m is a fraction, the method of superposition can not be used.
• A differential technique known as the S-curve method is adopted in such cases, and this method is applicable for rational values of m.

Construction of S-Curve-

By definition, an S-curve is obtained by adding a string of D-h unit hydrographs each lagged by D-hours from one another.

⇒  S(t) = U(t) + S (t - D)

Where,

S(t) = Ordinate of S-curve at time 't"

U(t) = Ordinate of Unit hydrograph at time "t"

Derivation of T-hour Unit Hydrograph-

Consider two D-h S-curves A and B displaced by T-h, if the ordinates of B are subtracted from that of A, the resulting curve is a DRH produced by a rainfall excess of duration T-h and magnitude T/D cm.

Hence if the ordinate differences of A and B (Sa-Sb) are divided by T/D, the resulting ordinates denote a hydrograph due to an ER of 1 cm and of duration T hour unit hydrograph.

Given data and Calculation:

 

For Obtaining S- Curve,

i) S(0) = U (0) + S (0 - 2) = 0

ii) S(1) = U (1) + S (1 - 2) = 2

iii) S(2) = U(2) + S(0) = 8 + 0 = 8

iv) S(3) = U(3) + S(3-2) = U(3) + S(1) = 10 + 2 = 12

Concept:

When the pipes are connected in parallel then head loss in both the pipes will be the same.

When the pipes are connected in series then the total head loss is the sum of the individual head losses in pipes.

Head loss due to friction in a pipe of length l is given as,

\(h_f = \frac{ 4. f. L. V^2}{d \times 2g}\)

Where f = Coefficient of friction and f'= Friction factor = 4f

L = Length of pipe, V = velocity of flow in pipe,

d= diameter of pipe, g= acceleration due to gravity

It can be expressed in terms of discharge also.

 \(\rm H = \frac{4.f.L \left( \frac{4Q}{ \pi d^2 }\right)^2}{d \times 2g} = \frac{4 \times 16Q^2 f}{ \pi^2 \times 2g} \left[ \frac{L}{d^5} \right]\)

where Q = Discharge in the pipe

Given data and Calculation:

Given that,

Diameter of larger pipe (d1) = 20 cm

Diameter of smaller pipe (d2) = 10 cm

The difference in head level of reservoir = 5 m

As the pipes are connected in parallel so the head loss in both the pipes will be the same.

h_f1 = h_f2

\(\rm \frac{8 Q_1^2}{\pi^2 g} \times \frac{fl}{D_1^5} = \frac{8 Q_2^2}{\pi^2g} \times \frac{fl}{D_2^5}\)

\(\left( \frac{Q_1}{Q_2} \right)^2 = \left( \frac{D_1}{D_2} \right)^5\)

\(\frac{Q_1}{Q_2} = \left( \frac{D_1}{D_2} \right)^{5/2} = \left( \frac{0.2}{0.1} \right)^{5/2}= 5.66\)

Concept:

Consider a frictionless horizontal channel of width B1 carrying a discharge Q at a depth y1 as shown in the figure.

At the Section 2 the channel width has been constricted to B2 by a smooth transition. Since there are no losses involved and since the bed elevations at Sections 1 and 2 are same, the specific energy at Section 1 is equal to the specific energy at the Section 2.

\(\rm E_1=y_1+\frac{V_1^2}{2g}=y_1+\frac{Q^2}{2gB_1^2y_2^2}\)

and

\(\rm E_2=y_2+\frac{V_2^2}{2g}=y_2+\frac{Q^2}{2gB_2^2y_2^2}\)

Let q = Flow per unit width at any section = Q/B

For subcritical flow, If B₂ is made smaller, then q₂ will increase and y₂ will decrease as shown in figure. The limit of the contracted width B₂ = B_2m is obviously reached when corresponding to E₁, the discharge intensity q₂ = q_m, i.e., the maximum discharge intensity for a given specific energy (critical-flow condition) will prevail.

At this minimum width, y₂ = critical depth at Section 2 (Y_cm)

For a rectangular channel, at critical flow \(\rm Y_c=\frac{2}{3}E_c\)

Since E₁ = E_cm

\(\rm y_2=y_{cm}=\frac{2}{3}E_{cm}=\frac{2}{3}E_1\) and 

\(\rm y_c=\left(\frac{Q^2}{B_{2m}^2g}\right)^{1/3}\) or \(\rm B_{2m}=\sqrt{\frac{Q^2}{g y^3_{cm}}}\)

Given data and Calculation-

Given that, Section is rectangular

Initial depth = 2 m, width = 3 m

discharge =12 m3/s, g = 9.8 m/s2 

Q = A₁ V₁

12 = 3 × 2 × V₁

V₁ = 2 m/s

\(\rm F_{r_1}=\frac{V_1}{\sqrt{gy_1}}=\frac{2}{\sqrt {g\times 2}}=0.45<1\)

As the Froud number is less than 1 then that is known as subcritical flow,

\(\rm (B_2)_{min}=\sqrt{\frac{27Q^2}{8gE_1^3}}\)

\(\rm E_1=y_1+\frac{V_1^2}{2g}=2+\frac{(2)^2}{2g}=2.2038\ m\)

\(\rm (B_2)_{min}=\sqrt{\frac{27\times12^2}{8\times9.81\times(2.2038)^3}}\)

= 2.15 m

Concept:

No. of moles = \(\frac{Given-Weight}{Molecular -weight}\)

If the chemical formula for any compound is given by A_xB_y, then it contains X moles of A and Y moles of B.

Given data and Calculation-

Given that,

Atomic weight of N = 14 g/mol, H = 1 g/mol, O = 16 g/mol

Concentration of Ammonia = 34 mg/liter and Nitrate = 6.2 mg/liter

Molecular weight of Ammonia = 14 + (3 × 1) = 17 gm

Molecular weight of Nitrate = 14 + (3 × 16) =62 gm

No. of moles of Ammonia = 0.034/17 = 0.002 

No. of moles of Nitrate = 0.0062/62 = 0.0001

1 mole of Ammonia and Nitrate contains 1 mole of Nitrogen.

So total number of moles of Nitrogen = 0.002 + 0.0001 = 0.0021 moles/liter

Concentration in mg/liter = 0.0021 x 1000 x 14 = 29.4 mg/liter

Concept-

The BOD test measures the strength of the wastewater by measuring the amount of oxygen used by the bacteria as they stabilize the organic matter under controlled conditions of time and temperature.

BOD_t = (initial Dissolved Oxygen - Dissolved oxygen after t days) x Dilution factor

Dilution factor = \(\frac{Total volume after dilution}{Initial volume}\)

An x% sewage sample means in the total volume of 100 unit initial volume of sample is x unit, so DF = 100/x.

L_t = L₀ e^-kt

Where BOD present   after any time t is given by L_t.

L₀ is the ultimate BOD.

K is deoxygeation constant 

BOD required for any tie t = L₀ - L_t

BOD_t = L₀ (1 - e^-kt)

Given data and Calculation-

Depletion of Dissolved oxygen = 10 mg/L

Dilution factor = 100/2 =50

BOD₃ = 10 × 50=500 mg/L

L₃ = L₀ (1 - e^{-0.23 × 3})

⇒ 500 = L0 (1 - e-0.23 × 3)

⇒ L₀ = \(\frac{500}{(1-e^{-0.23\times3})}\) = 1003.162 = 1003 mg/L

Concept:

The main principle involved in the sedimentation tank is to reduce the flow velocity of water which allows the major amount of suspended particles to settle down.

The velocity with which the particle is settling is known as settling velocity.

The volume of water applied per unit time per unit horizontal surface area is called overflow velocity.

It is also called surface loading rate or surface overflow rate.

Overflow velocity must be less than the settling velocity of the particle otherwise the particle or sediment will escape without settling.

V_o = \(\frac{Discharge}{Surface- Area}\)

Surface Area = Width x Length

Flow through velocity = \(\frac{Discharge}{flow- Area}\)

The time taken by water to travel from inlet to outlet is termed as detention period or detention time.

T_d = \(\frac{Volume- of- tank}{Discharge}\)

In sedimentation tanks, efficiency purely depends upon the settling velocity and overflow velocity.

Fraction of particles removed = \(\frac{Settling- Velocity -for- the -Particle}{Overlfow-Velocity}\)

Given data and Calculation-

Given that, Depth = H = 3 m, Width = B = 5 m, Length = L = 40 m

Discharge = 500 m3/h

settling velocity of particle = 1.0 m/h

Surface Area = 40 × 5 =200 m²

Overflow velocity = \(\frac{Discharge}{Surface- Area}\) = 500/200 = 2.5 m/h

Fraction = 1/2.5 = 0.4

Concept:

Effective green time-

Effective green time is the actual time available for the vehicles to cross the intersection.

It is the sum of actual green time (G_i) plus the yellow minus the applicable lost times. This lost time is the sum of start-up lost time (l₁) and clearance lost time (l₂) denoted as t_L.

Thus effective green time can be written as ⇒ g_i = G_i + Y_i - t_L

Cycle time is the time taken by a signal to complete one full cycle of iterations. i.e. one complete rotation through all signal indications. It is denoted by C.

C = G_i + Y_i +R_i

Where G_i, R_i, Y_i =  Green time, Red time, yellow time for 'i'th phase respectively.

Given data and Calculation-

Cycle time = 100 s

Amber time = 4 s, Red time = 50 s, Total lost time = 2 s

C - (Yi +Ri) = Gi 

Green time = 100 - (50 + 4) = 46 s

gi = Gi + Yi - tL

Effective green time = 46 + 4 - 2 = 48 s

Concept-

At a traffic intersection arrival of vehicles follows a Poisson distribution.

The Poisson’s model that accounts for nonuniformity of flow which is derived by assuming the random pattern of vehicle arrivals at a specified point is given by

\(P\left( n \right) = \frac{{{{\left( {λ t} \right)}^n} \times {e^{ - λ t}}}}{{n!}}\)  ..... ( 1 )

where

P(n) = probability of having n vehicles arrive in time t,

n - Number of vehicles in time t

λ = average vehicle flow or arrival rate in vehicles per unit time,

t = time interval

e = base of the natural logarithm (e = 2.718)

Time headway is defined as the time difference between any two successive vehicles when they cross a given point.

Given data and Calculation:

• Arrival rate of car = 4 veh/hr
• Arrival rate of bus = 2 veh/hr
• Arrival rate of vehicles = 4 + 2 = 6 veh/hour = 3 veh/30 min

The probability of observing at least 2 vehicles in 30 minutes is,

P (x ≥ 2) = 1 - P (x ≤ 1)

= 1 - [P (x = 0) + P (x = 1)]

= \(1-\left[\frac{e^{-\lambda}\lambda^0}{0!}+\frac{e^{-\lambda}\lambda^1}{1!}\right]\)

\(1-\left[\frac{e^{-3}}{1}+\frac{e^{-3}\times3}{1}\right]\)

\(=1-\frac{4}{e^3}\)

= 0.8

Ans-0.8

Concept-

.The peak-hour factor (PHF) is descriptive of trip generation patterns and may apply to an area or portion of a street and highway system. The PHF is typically calculated from traffic counts.

It is the average volume during the peak 60 minute period V₆₀ divided by (60/n) times the average volume during the peak “n” minute’s period V_n.

PHF = V₆₀/(60 × Vn /n)

Where V_n is the peak n minute flow.

Given data and Calculation-

 

Total number of vehicles in 60 minutes = 10 + 11 + 12 +15 + 13 + 11 = 72 vehicles

Peak 10 minutes flow = 15 vehicles

PHF = \(\frac{72}{15 ( \frac{60}{10})}\)=72/90  = 0.8

Concept-

ESWL(Equivalent Single Wheel Load)-

To carry the maximum load within the specified limit and to carry a greater load, dual wheel, or dual tandem assembly is often used.

Equivalent single wheel load (ESWL) is the single wheel load having the same contact pressure, which produces a same value of maximum stress, deflection, tensile stress or contact pressure at the desired depth.

The procedure of finding the ESWL for equal stress criteria is a semi-rational method, known as Boyd and Foster method, based on the following assumptions:

• The contact area is circular.
• The influence angle is 45 degrees.
• Soil medium is elastic, homogeneous, and isotropic half-space.

 In a dual wheel assembly, let d is a clear gap between the wheels,

 S is the spacing between the centers of the wheels

 ESWL is given by

 \({\log _{10}}ESWL = {\log _{10}}P + \frac{{0.301 \times {{\log }_{10}}\left( {\frac{Z}{{d/2}}} \right);;}}{{{{\log }_{10}}\left( {\frac{{2S}}{{d/2}}} \right)}}\)  --- (1)

Where,

ESWL is an equivalent single-wheel load.

P is wheel load, Z is desired depth

Up to depth of d/2, no stress overlap occurs

∴ ESWL = P

And for depth greater than or equal to 2S, the effect of overlap is such that we can consider

ESWL = 2 × P

Given data and Calculation-

The ratio of Equivalent Single Wheel Load (ESWL) at depth z = d/2 to the ESWL at depth z = 2s has to be found.

At depth d/2, the ESWL is P and at depth 2s, the ESWL is 2P.

Ratio = P/2P = 0.5